Edward is making a rectangular picture frame. He wants the perimeter of the frame to be no more than 96 inches. He also wants the length of the frame to be greater than or equal to the square of 4 inches less than its width.

Create a system of inequalities to model the above situation and use it to determine how many of the solutions are viable.

A.) Part of the solution region includes a negative length; therefore, all solutions are not viable for the given situation.

B.) No part of the solution region is viable because the length or width cannot be negative.

C.) The entire solution region is viable.

D.) Part of the solution region includes a negative width; therefore, not all solutions are viable for the given situation.

Part I.
Let x represent the width of the frame, any y represent the length. The sum of length and width will be half the perimeter, so one inequality is
x + y ≤ 48
If we assume the length restriction applies to the numeric values measured in inches, then the second inequality can be written as
y ≥ (x-4)²

Part II.
The solution is shown in the figure where the colored regions overlap. Unless we also add restrictions x>0, y>0, some of the solution space is located where x < 0. Thus it is appropriate to conclude ...
D.) Part of the solution region includes a negative width; therefore, not all solutions are viable for the given situation.

kjuli1766 kjuli1766 · Answer 2 · 2022-05-15T05:17:22+02:00

The correct option is D.) Part of the solution region includes a negative width; therefore, not all solutions are viable for the given situation.

What is perimeter?

The total sum of the four sides of the quadrilateral is called perimeter.

For a rectangle perimeter is 2(length + breadth)

How to find the inequalities?

Let the length of the frame be x inches and the width of the frame be y inches

Here, the perimeter of the frame to be no more than 96 inches.

∴According to the problem,

2 (x+y) ≤ 96

⇒(x+y) ≤ 48..........(1)

Again, length of the frame to be greater than or equal to the square of 4 inches less than its width

∴According to the problem,

x ≥ [tex](y-4)^{2}[/tex] ....................(2)

Combining these inequalities,

Since, x and y are length and breadth, only positive values will be considered.
x, y ≥ 0.

Clearly, the solution region includes a negative part of y axis which is actually the width of the frame.

Option A states that a part of the solution region includes a negative length, which is wrong.

Option B states that no part of the solution region is viable, which is also wrong because all the solutions in the first quadrant will be viable since in the first quadrant x, y ≥ 0.

Option C states that the entire solution region is viable which is also wrong because the solution region includes a negative part of y axis which is not viable

∴ Option D is correct.

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